Problem: Find the greatest constant $M,$ so that
\[\frac{a^2 + b^2}{c^2} > M\]whenever $a,$ $b,$ and $c$ are the sides of a triangle.
Explanation: Consider a triangle $ABC$ where $a = b.$

[asy]
unitsize (3 cm);

pair A, B, C;

A = (0,0);
B = (2,0);
C = (1,0.2);

draw(A--B--C--cycle);

label("$A$", A, W);
label("$B$", B, E);
label("$C$", C, N);
label("$a$", (B + C)/2, N);
label("$a$", (A + C)/2, N);
label("$c$", (A + B)/2, S);
[/asy]

As $\angle ACB$ approaches $180^\circ,$ $c$ approaches $2a,$ so $\frac{a^2 + b^2}{c^2}$ approaches $\frac{a^2 + a^2}{(2a)^2} = \frac{1}{2}.$  This means $M \le \frac{1}{2}.$

On the other hand, by the triangle inequality, $c < a + b,$ so
\[c^2 < (a + b)^2 = a^2 + 2ab + b^2.\]By AM-GM, $2ab < a^2 + b^2,$ so
\[c^2 < 2a^2 + 2b^2.\]Hence,
\[\frac{a^2 + b^2}{c^2} > \frac{1}{2}.\]Therefore, the largest such constant $M$ is $\boxed{\frac{1}{2}}.$